S.M. Yusuf Mathematical Methods Chapter 8 Infinite Series Solved Exercises & Convergence Tests PDF Guide | PU, UOS, QAU, FBISE

Theoretical Foundations & Core Theorems

In real analysis and advanced calculus, an infinite series is defined as the formal sum of an infinite sequence of real numbers $(a_n)_{n=1}^{\infty}$, denoted as:

$$\sum_{n=1}^{\infty} a_n = a_1 + a_2 + a_3 + \dots + a_n + \dots$$

The convergence or divergence of an infinite series is intrinsically tied to the behavior of its sequence of partial sums $(S_k)$, where $S_k = \sum_{n=1}^{k} a_n$. If the limit $\lim_{k \to \infty} S_k = S$ exists as a finite real number, the series is said to converge to the sum $S$. Otherwise, the series diverges.

To analyze infinite series in degree-level examinations (PU, UOS, QAU, KU), students must master a systematic sequence of tests described in Chapter 8 of S.M. Yusuf's Mathematical Methods:

• The $n$-th Term Test for Divergence: A fundamental necessary condition for convergence. If $\lim_{n \to \infty} a_n \neq 0$ (or if the limit does not exist), then $\sum a_n$ diverges. Warning: $\lim_{n \to \infty} a_n = 0$ does not imply convergence (e.g., the harmonic series $\sum \frac{1}{n}$).
• Integral Test (Cauchy Integral Criterion): If $f(x)$ is continuous, positive, and strictly decreasing on $[1, \infty)$ such that $f(n) = a_n$, then $\sum_{n=1}^{\infty} a_n$ and $\int_{1}^{\infty} f(x) dx$ either both converge or both diverge.
• Basic Comparison Test (BCT) & Limit Comparison Test (LCT): Used primarily for positive term series by comparing $a_n$ to a known benchmark series $b_n$ (such as $p$-series $\sum \frac{1}{n^p}$ or geometric series $\sum r^n$). In LCT, if $\lim_{n \to \infty} \frac{a_n}{b_n} = L$, where $0 < L < \infty$, then both series behave identically.
• D'Alembert's Ratio Test: Evaluates $L = \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right|$. The series converges absolutely if $L < 1$, diverges if $L > 1$, and the test is inconclusive if $L = 1$. Excellent for factorials ($n!$) and powers ($a^n$).
• Cauchy's Root Test: Evaluates $L = \lim_{n \to \infty} \sqrt[n]{|a_n|}$. Converges if $L < 1$, diverges if $L > 1$, inconclusive if $L = 1$. Highly effective for expressions raised to the $n$-th power.
• Raabe's Test: Utilized specifically when D'Alembert's Ratio Test fails ($L = 1$). Computes $L = \lim_{n \to \infty} n \left( \frac{a_n}{a_{n+1}} - 1 \right)$. Converges if $L > 1$, diverges if $L < 1$, and fails if $L = 1$.
• Leibniz Test for Alternating Series: An alternating series $\sum (-1)^{n+1} a_n$ (where $a_n > 0$) converges if two conditions are met:
  1. $a_{n+1} \le a_n$ for all $n \ge N$ (monotonically non-increasing).
  2. $\lim_{n \to \infty} a_n = 0$.
• Absolute vs. Conditional Convergence: A series $\sum a_n$ is absolutely convergent if $\sum |a_n|$ converges. If $\sum a_n$ converges but $\sum |a_n|$ diverges, the series is conditionally convergent.

Formula Summary & Quick Reference

Test Name	Mathematical Formula / Condition	Convergence Criteria	Divergence / Failure Criteria	Primary Application
$n$-th Term Test	$$\lim_{n \to \infty} a_n$$	Inconclusive if limit is $0$	Diverges if limit $\neq 0$ or DNE	First line of defense for any series
Geometric Series	$$\sum_{n=0}^{\infty} a r^n$$	Converges if $|r| < 1$ (Sum = $\frac{a}{1-r}$)	Diverges if $|r| \ge 1$	Exponential terms
$p$-Series Test	$$\sum_{n=1}^{\infty} \frac{1}{n^p}$$	Converges if $p > 1$	Diverges if $p \le 1$	Polynomial / Rational power benchmarks
Limit Comparison (LCT)	$$L = \lim_{n \to \infty} \frac{a_n}{b_n}$$	Both behave identically if $0 < L < \infty$	If $L=0$ and $\sum b_n$ div, inconclusive	Algebraic fractions of polynomials/roots
D'Alembert Ratio Test	$$L = \lim_{n \to \infty} \left|\frac{a_{n+1}}{a_n}\right|$$	Converges if $L < 1$	Diverges if $L > 1$; Inconclusive if $L = 1$	Factorials $n!$, exponents $k^n$
Cauchy Root Test	$$L = \lim_{n \to \infty} |a_n|^{1/n}$$	Converges if $L < 1$	Diverges if $L > 1$; Inconclusive if $L = 1$	Expressions wrapped in $n$-th powers $(f(n))^n$
Raabe's Test	$$L = \lim_{n \to \infty} n \left( \frac{a_n}{a_{n+1}} - 1 \right)$$	Converges if $L > 1$	Diverges if $L < 1$; Inconclusive if $L = 1$	Failsafe when Ratio Test yields $L = 1$
Integral Test	$$\int_{1}^{\infty} f(x) dx$$	Converges if integral evaluates to finite value	Diverges if integral approaches $\pm \infty$	Logarithmic terms, inverse trig functions
Leibniz Test	$$\sum (-1)^{n-1} a_n$$	Converges if $a_{n+1} \le a_n$ and $\lim a_n = 0$	Test fails if $\lim a_n \neq 0$ (Diverges by $n$-th term)	Alternating sign series

Step-by-Step Solved Board Exam Questions

Problem 1 (S.M. Yusuf Ex 8.2 - Limit Comparison Test)

Question: Test the convergence of the infinite series:

$$\sum_{n=1}^{\infty} \frac{\sqrt{n}}{n^2 + 1}$$

Solution:

Step 1: Identify the general term $a_n$.

$$a_n = \frac{\sqrt{n}}{n^2 + 1}$$

Observe that $a_n > 0$ for all $n \ge 1$. Hence, this is a positive-term series.

Step 2: Choose an appropriate comparison series $b_n$.
To determine $b_n$, extract the highest powers of $n$ from the numerator and denominator of $a_n$:

$$b_n = \frac{n^{1/2}}{n^2} = \frac{1}{n^{2 - 1/2}} = \frac{1}{n^{3/2}}$$

Step 3: Apply the Limit Comparison Test (LCT).
Evaluate the limit $L = \lim_{n \to \infty} \frac{a_n}{b_n}$:

$$L = \lim_{n \to \infty} \left( \frac{\frac{\sqrt{n}}{n^2 + 1}}{\frac{1}{n^{3/2}}} \right) = \lim_{n \to \infty} \left( \frac{\sqrt{n} \cdot n^{3/2}}{n^2 + 1} \right) = \lim_{n \to \infty} \frac{n^2}{n^2 + 1}$$

Divide the numerator and denominator by $n^2$:

$$L = \lim_{n \to \infty} \frac{1}{1 + \frac{1}{n^2}} = \frac{1}{1 + 0} = 1$$

Step 4: Conclusion based on LCT and $p$-series test.
Since $L = 1$ is a finite, non-zero real number ($0 < L < \infty$), both series $\sum a_n$ and $\sum b_n$ behave identically.
Now examine $\sum b_n = \sum_{n=1}^{\infty} \frac{1}{n^{3/2}}$. This is a $p$-series with $p = \frac{3}{2}$.
Since $p = \frac{3}{2} > 1$, the auxiliary series $\sum b_n$ converges.
Therefore, by the Limit Comparison Test, the original series $\sum_{n=1}^{\infty} \frac{\sqrt{n}}{n^2 + 1}$ is convergent.

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Problem 2 (S.M. Yusuf Ex 8.3 - D'Alembert Ratio Test)

Question: Test the convergence of the series:

$$\sum_{n=1}^{\infty} \frac{n! \, 2^n}{n^n}$$

Solution:

Step 1: Write down $a_n$ and $a_{n+1}$.

$$a_n = \frac{n! \, 2^n}{n^n}$$ $$a_{n+1} = \frac{(n+1)! \, 2^{n+1}}{(n+1)^{n+1}}$$

Step 2: Form the ratio $\frac{a_{n+1}}{a_n}$.

$$\frac{a_{n+1}}{a_n} = \frac{(n+1)! \, 2^{n+1}}{(n+1)^{n+1}} \times \frac{n^n}{n! \, 2^n}$$

Simplify factorials and exponential powers:

$$(n+1)! = (n+1) \cdot n!, \quad 2^{n+1} = 2 \cdot 2^n, \quad (n+1)^{n+1} = (n+1)(n+1)^n$$

Substitute these back into the expression:

$$\frac{a_{n+1}}{a_n} = \frac{(n+1) n! \cdot 2 \cdot 2^n}{(n+1)(n+1)^n} \times \frac{n^n}{n! \cdot 2^n}$$

Cancel out common terms $n!$, $2^n$, and $(n+1)$:

$$\frac{a_{n+1}}{a_n} = \frac{2 \cdot n^n}{(n+1)^n} = 2 \left( \frac{n}{n+1} \right)^n = \frac{2}{\left( \frac{n+1}{n} \right)^n} = \frac{2}{\left( 1 + \frac{1}{n} \right)^n}$$

Step 3: Calculate the limit $L = \lim_{n \to \infty} \frac{a_{n+1}}{a_n}$.

$$L = \lim_{n \to \infty} \frac{2}{\left( 1 + \frac{1}{n} \right)^n}$$

Using the standard calculus limit identity $\lim_{n \to \infty} \left( 1 + \frac{1}{n} \right)^n = e$:

$$L = \frac{2}{e}$$

Step 4: Apply D'Alembert's Ratio Test rule.
Since $e \approx 2.71828$, we have $L = \frac{2}{2.71828} < 1$.
Because $L < 1$, by D'Alembert's Ratio Test, the series $\sum_{n=1}^{\infty} \frac{n! \, 2^n}{n^n}$ converges.

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Problem 3 (S.M. Yusuf Ex 8.4 - Failure of Ratio Test & Application of Raabe's Test)

Question: Test the convergence of the series:

$$\sum_{n=1}^{\infty} \left( \frac{1 \cdot 3 \cdot 5 \cdots (2n-1)}{2 \cdot 4 \cdot 6 \cdots (2n)} \right)^2$$

Solution:

Step 1: Define $a_n$ and $a_{n+1}$.

$$a_n = \left( \frac{1 \cdot 3 \cdot 5 \cdots (2n-1)}{2 \cdot 4 \cdot 6 \cdots (2n)} \right)^2$$ $$a_{n+1} = \left( \frac{1 \cdot 3 \cdot 5 \cdots (2n-1)(2n+1)}{2 \cdot 4 \cdot 6 \cdots (2n)(2n+2)} \right)^2$$

Step 2: Try D'Alembert's Ratio Test first.

$$\frac{a_{n+1}}{a_n} = \left( \frac{2n+1}{2n+2} \right)^2$$ $$L = \lim_{n \to \infty} \frac{a_{n+1}}{a_n} = \lim_{n \to \infty} \left( \frac{2 + \frac{1}{n}}{2 + \frac{2}{n}} \right)^2 = \left( \frac{2}{2} \right)^2 = 1$$

Since $L = 1$, D'Alembert's Ratio Test fails (inconclusive). We must apply Raabe's Test.

Step 3: Setup Raabe's Test formula.
Compute $L_R = \lim_{n \to \infty} n \left( \frac{a_n}{a_{n+1}} - 1 \right)$:

$$\frac{a_n}{a_{n+1}} = \left( \frac{2n+2}{2n+1} \right)^2 = \frac{4n^2 + 8n + 4}{4n^2 + 4n + 1}$$

Subtract $1$:

$$\frac{a_n}{a_{n+1}} - 1 = \frac{4n^2 + 8n + 4 - (4n^2 + 4n + 1)}{4n^2 + 4n + 1} = \frac{4n + 3}{4n^2 + 4n + 1}$$

Multiply by $n$ and evaluate the limit:

$$L_R = \lim_{n \to \infty} n \left( \frac{4n + 3}{4n^2 + 4n + 1} \right) = \lim_{n \to \infty} \frac{4n^2 + 3n}{4n^2 + 4n + 1}$$

Divide numerator and denominator by $n^2$:

$$L_R = \lim_{n \to \infty} \frac{4 + \frac{3}{n}}{4 + \frac{4}{n} + \frac{1}{n^2}} = \frac{4 + 0}{4 + 0 + 0} = 1$$

Notice: Raabe's test yields $1$ here! Let's re-verify the degree carefully. When $L_R = 1$, Raabe's Test is inconclusive. Let's use Higher-Order / Gauss's Test or re-evaluate the expansion:

$$\frac{a_n}{a_{n+1}} = \left(1 + \frac{1}{2n+1}\right)^2 = 1 + \frac{2}{2n+1} + \frac{1}{(2n+1)^2} = 1 + \frac{1}{n} \left(\frac{2n}{2n+1}\right) + O(1/n^2) = 1 + \frac{1}{n} + O(1/n^2)$$

Since the coefficient of $\frac{1}{n}$ in Gauss's expansion is $\lambda = 1 \le 1$, the series diverges.

By Gauss's Criterion (or refined Raabe analysis), the given series diverges.

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Problem 4 (S.M. Yusuf Ex 8.5 - Absolute vs Conditional Convergence)

Question: Determine whether the alternating series converges absolutely, conditionally, or diverges:

$$\sum_{n=1}^{\infty} (-1)^{n+1} \frac{1}{\sqrt{n}}$$

Solution:

Step 1: Test the original series for simple convergence using the Leibniz Test.
The series is $\sum_{n=1}^{\infty} (-1)^{n+1} a_n$, where $a_n = \frac{1}{\sqrt{n}} > 0$.

Check the two Leibniz conditions:

1. Monotonic Decrease: $\sqrt{n+1} > \sqrt{n} \implies \frac{1}{\sqrt{n+1}} < \frac{1}{\sqrt{n}} \implies a_{n+1} < a_n$ for all $n \ge 1$. Condition satisfied.
2. Limit to Zero: $\lim_{n \to \infty} a_n = \lim_{n \to \infty} \frac{1}{\sqrt{n}} = 0$. Condition satisfied.

Since both conditions hold, the series $\sum_{n=1}^{\infty} (-1)^{n+1} \frac{1}{\sqrt{n}}$ converges by the Leibniz Alternating Series Test.

Step 2: Test for Absolute Convergence.
Consider the series of absolute values $\sum_{n=1}^{\infty} |(-1)^{n+1} \frac{1}{\sqrt{n}}| = \sum_{n=1}^{\infty} \frac{1}{n^{1/2}}$.

This is a $p$-series with $p = \frac{1}{2}$.
Since $p = \frac{1}{2} \le 1$, the absolute series $\sum_{n=1}^{\infty} \frac{1}{\sqrt{n}}$ diverges.

Step 3: Final Classification.
Because the original alternating series converges, but the series of absolute values diverges, the series $\sum_{n=1}^{\infty} (-1)^{n+1} \frac{1}{\sqrt{n}}$ is conditionally convergent.

Interactive Practice Quiz (Clickable MCQs)

Frequently Asked Questions (FAQs)

Q1: Why does $\lim_{n \to \infty} a_n = 0$ NOT guarantee that the series $\sum a_n$ converges?
Answer: The condition $\lim_{n \to \infty} a_n = 0$ is a necessary condition for convergence, but not a sufficient one. It simply means the terms are shrinking to zero, but it does not tell us whether they shrink fast enough for the partial sums to approach a finite limit. The classic counterexample is the harmonic series $\sum_{n=1}^{\infty} \frac{1}{n}$, where $\lim_{n \to \infty} \frac{1}{n} = 0$, yet the series diverges to infinity.

Q2: What should I do when D'Alembert's Ratio Test yields $L = 1$ in university exams?
Answer: When $L = 1$, D'Alembert's Ratio Test provides no information. For Pakistani university syllabus (S.M. Yusuf Ch 8), follow this decision hierarchy:

1. If the general term involves factorials or products of double odds/evens (e.g., $1 \cdot 3 \cdot 5 \dots$), switch to Raabe's Test or Gauss's Test.
2. If the general term is purely algebraic/polynomial without factorials, use the Limit Comparison Test (LCT) with a $p$-series.
3. If the term involves logarithms, apply the Cauchy Integral Test.

Q3: What is the practical difference between Absolute Convergence and Conditional Convergence?
Answer:

• Absolute Convergence: Means $\sum |a_n|$ converges. Absolutely convergent series are extremely well-behaved; you can rearrange their terms in any order without changing the sum (Riemann Series Theorem).
• Conditional Convergence: Means $\sum a_n$ converges, but $\sum |a_n|$ diverges. These series are sensitive to term order—rearranging their terms can change the final sum to any desired real value or cause the series to diverge.